Given data:
* The mass of the car is m = 1200 kg.
* The distance up to which Tyler can see is d = 15 m.
* The coefficient of friction between the road and the car is,
[tex]\mu=0.67[/tex]
* The force applied by Tyler on the car is,
[tex]F=1.3\times10^4\text{ N}[/tex]
Solution:
The diagrammatic representation of the given case is,
(a). The weight of the car is,
[tex]W_{fg}=mg[/tex]
where g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} W_{fg}=1200\times9.8 \\ W_{fg}=11760\text{ N} \end{gathered}[/tex]
Thus, the weight of the car is 11760 N.
(b). The normal force acting on the car is,
[tex]\begin{gathered} N=W_{fg} \\ N=11760\text{ N} \end{gathered}[/tex]
The frictional force acting on the car is,
[tex]\begin{gathered} F_r=\mu N \\ F_r=0.67\times11760 \\ F_r=7879.2\text{ N} \end{gathered}[/tex]
Thus, the net force acting on the car is,
[tex]\begin{gathered} W_{\text{net}}=F-F_r \\ W_{\text{net}}=1.3\times10^4-7879.2 \\ W_{\text{net}}=13000-7879.2 \\ W_{\text{net}}=5120.8\text{ N} \end{gathered}[/tex]
Thus, the net force acting on the car is 5120.8 N.
(c). According to Newton's second law, the acceleration of the car is,
[tex]\begin{gathered} W_{\text{net}}=ma \\ a=\frac{W_{\text{net}}}{m} \end{gathered}[/tex]
Substituting the known values,
[tex]\begin{gathered} a=\frac{5120.8}{1200} \\ a=4.27ms^{-2} \end{gathered}[/tex]
Thus, the acceleration of the car is 4.27 meters per second squared.
(d). The initial velocity of the car is zero, by the kinematics equation, the time taken by the car to reach the gas station is,
[tex]d=ut+\frac{1}{2}at^2[/tex]
where u is the initial velocity,
Substituting the known values,
[tex]\begin{gathered} 15=0+\frac{1}{2}\times4.27\times t^2 \\ 15=2.14t^2 \\ t^2=\frac{15}{2.14} \\ t^2=7 \end{gathered}[/tex]
By taking the square root,
[tex]t=2.6\text{ s}[/tex]
Thus, the time taken by Tyler to reach the gas station is 2.6 seconds.
